Remark 2.1 \({\rm (i)}\)  The twisted de Rham cohomology groups \(H^*(M, H)\)  \((*=o,e)\)  depend on the closed form \(H\)  and not just on its cohomology
class. If \(H\)  and \(H^{^{\prime }}\)  are cohomologous, then \(H^*(M, H)\cong H^*(M, H^{^{\prime }})\)  \((\) see [1]}\()\) .
